Lorentz force and work done: where is the glitch?

1. Aug 15, 2014

pleco

Where is the glitch?

1. Let a single free electron move along x axis with constant speed Vx. It encounters magnetic field on its right side, so the Lorentz force accelerates it along y axis and it gains speed Vy.
--- Glitch or pass?

2. The electron does not slow down in x direction when it starts accelerating in y direction, but instead this new gained speed in y direction adds up to its Vx velocity vector, increases its speed and kinetic energy along with changing its trajectory.
--- Glitch or pass?

x direction of wire1 electrons drift
y direction of wire2 B field
z direction of Lorentz force acting on wire1 electrons

Direction of wire1 electrons displacement due to Lorentz force is in z direction downwards, the same direction Lorentz force is pointing to.
--- Glitch or pass?

Please mark your axis and be specific about your directions. That vdt is wrong velocity vector, electrons drift vector. That is not the displacement in the direction of the force. Acceleration, velocity and the displacement due to Lorentz force are all vectors pointing where the Lorentz force is pointing itself, that's given by F= ma.

{0,0,1} dot {1,0,0} = 0 <- drift velocity vector = wrong displacement vector

{0,0,1} dot {0,0,1} = 1 <- acceleration/displacement vector = correct vector

Last edited: Aug 15, 2014
2. Aug 15, 2014

Staff: Mentor

There is a glitch in 1 and 2. The electron will gain velocity in y but lose velocity in x such that it's speed will be constant.

I can post the math later.

3. Aug 15, 2014

milesyoung

Both are false, any change in velocity in the y-direction will come with a change in the velocity in the x-direction such that the magnitude of the velocity vector of the electron stays constant.

If the net force on an object and its velocity vector are always orthogonal then THE OBJECT WILL NOT CHANGE ITS SPEED. This is basic mechanics.

With regards to Ampère's force law - you bring it up every time, but to understand why it's NOT a counterexample, you first have to understand why the magnetic force can't do any work on a point charge. Now please, just show me where the math is wrong here:

We have an electron traveling in some magnetic field $\mathbf{B}$ (which doesn't have to be uniform) with the instantaneous velocity $\mathbf{v}$. This electron is not one of the many free electrons in a conductor. This electron is simply traveling in space, unconstrained but affected by the magnetic field.

So let's calculate the work done by the magnetic force on the electron from some time $t_1$ to $t_2$:
$$W = \int_{t_1}^{t_2} \mathbf{F_\mathbf{mag}} \cdot \mathbf{v} \, \mathrm{d}t = \int_{t_1}^{t_2} \left( q \mathbf{v} \times \mathbf{B} \right) \cdot \mathbf{v} \, \mathrm{d}t = \int_{t_1}^{t_2} 0 \, \mathrm{d}t = 0$$
Now, please don't bring up Ampère's force law yet again - it's not a counterexample even though you think it is. Just stick to the magnetic force component of the Lorentz force law for now. As you say, it's always right.

4. Aug 15, 2014

pleco

I don't think that's how vector addition works. Let electron move with 10m/s along x-axis, then add 5m/s along y-axis: {10,0,0} + {0,5,0} = {10,5,0}. Vector magnitude increases from 10m/s to 11.18m/s.

You didn't mark axis or specify any directions. Please point at which step do you disagree, look:

Electron velocity along x-axis {1, 0, 0}
B_field vector along y-axis {0, 1, 0}
Force_vector = {1, 0, 0} cross {0, 1, 0} = {0, 0, 1}

So it means, the force vector is along z-axis, perpendicular to both the electron's initial velocity vector and B field vector. The electron feels this acceleration from Lorentz force and increases its z-axis velocity component. Thus, displacement vector = {0, 0, 1}, and therefore:

F dot v = {0, 0, 1} dot {0, 0, 1} = 1.

It's zero only if the two vectors are perpendicular, it is not zero if they are in the same direction, as they are in the this case with Lorentz force. Ok?

5. Aug 15, 2014

Staff: Mentor

$\left( q \mathbf{v} \times \mathbf{B} \right) \cdot \mathbf{v}=0$ for any vectors $v$ and $B$ and any value of $q$ no matter what axes and directions you choose.

6. Aug 15, 2014

milesyoung

You assume the velocity in the x-direction stays constant. It does not. The fact that you don't know this from whatever course you've had in introductory mechanics makes me think we should be talking about something much more basic.

Regardless of how you set it up, the magnetic force on the electron and its velocity vector will always be orthogonal. So as shown, the magnetic force can't do any work.

This is fine.

No, you don't get to just arbitrarily choose what you substitute into $\mathbf{F_\mathbf{mag}} \cdot \mathbf{v}$. You use the instantaneous magnetic force $\mathbf{F_\mathbf{mag}}$ and the instantaneous velocity $\mathbf{v}$:

F dot v = {0, 0, 1} dot {1, 0, 0} = 0

You need to brush up on your basic mechanics, e.g.:
This is just plain wrong. Like haruspex suggested, look up how uniform circular motion works.

If you want to discuss mechanics, then we can do that, but we're not going to have a discussion about EM fields etc.

Now I really have to get some sleep.

7. Aug 15, 2014

pleco

drift direction X-axis: {1, 0, 0}
field vector Y-axis: {0, 1, 0}
displacement Z-axis: {0, 0, 1}

({1, 0, 0} cross {0, 1, 0}) dot {0, 0, 1} = 1

8. Aug 15, 2014

rubi

I don't get why you think that $(q\vec{v}\times\vec{B})\cdot\vec{v} \neq 0$, but here is an elementary proof that it is indeed $=0$:

$$\vec{v} = \begin{pmatrix}v_x \\ v_y \\ v_z \end{pmatrix} \qquad\vec{B} = \begin{pmatrix}B_x \\ B_y \\ B_z \end{pmatrix}$$
$$\vec{F} = q\, \vec{v}\times\vec{B} = q\,\begin{pmatrix}v_x \\ v_y \\ v_z \end{pmatrix}\times\begin{pmatrix}B_x \\ B_y \\ B_z \end{pmatrix} = q\, \begin{pmatrix}v_y B_z - v_z B_y \\ v_z B_x - v_x B_z \\ v_x B_y - v_y B_x \end{pmatrix}$$
$$\vec{F}\cdot\vec{v} = q \, \begin{pmatrix}v_y B_z - v_z B_y \\ v_z B_x - v_x B_z \\ v_x B_y - v_y B_x \end{pmatrix} \cdot \begin{pmatrix}v_x \\ v_y \\ v_z \end{pmatrix} = q\,(v_y B_z - v_z B_y) v_x + q\,(v_z B_x - v_x B_z) v_y + q\,(v_x B_y - v_y B_x) v_z = 0$$
(Every positive term has a corresponding negative term, so everything cancels.)

9. Aug 15, 2014

Staff: Mentor

Here is the math I promised earlier

The force is $\mathbf{F}=q\mathbf{v}\times \mathbf{B} = m \mathbf{a} = m \frac{d}{dt}\mathbf{v}$. So the differential equation we are solving is:

$\frac{d}{dt}\mathbf{v}=\frac{q}{m}\mathbf{v}\times \mathbf{B}$

If you plug in $\mathbf{B}=(0,0,B)$ and the initial condition $\mathbf{v}(0)=(v_0,0,0)$ and you solve that differential equation then you get:

$\mathbf{v}=(v_0 \cos(\frac{q}{m} B t), -v_0 \sin(\frac{q}{m} B t),0)$

That is true, but not relevant. This problem is a vector differential equation, not a vector addition problem.

Last edited: Aug 15, 2014
10. Aug 15, 2014

pleco

Please provide some reference for that.

I drift direction X-axis: {1, 0, 0}
B field vector Y-axis: {0, 1, 0}
F force vector Z-axis: {0, 0, 1}

1. Lorentz force acts along Z-axis?

2. Lorentz force accelerate electrons along Z-axis?

It's not arbitrary, I chose Lorentz force and wire displacement direction. Why do you choose electrons drift direction as "displacement vector"? That's not the direction of wires displacement nor of Lorentz force. Are you not supposed to find out if there is any displacement in the direction of the force?

Last edited: Aug 15, 2014
11. Aug 15, 2014

pleco

Because that second velocity vector is supposed to be in the direction of the force and/or wire displacement. Why would you calculate work done relative to the direction of the current when neither the force nor wire displacement is in that direction? That equation works for electric force, but magnetic force is different and so we have to be more careful about what is it we are actually trying to calculate and chose appropriate vectors.

12. Aug 15, 2014

rubi

No, you can't randomly replace the quantities in the formula by other quantities. The formula for the work at an instant of time is $\mathrm d W=\vec{F}\cdot\mathrm d\vec{x}$, where $\vec{F} = q\,\vec{v}\times\vec{B}$ is the Lorentz force at that instant of time and $\mathrm d\vec{x}=\vec{v}\,\mathrm d t$ is the displacement at that instant of time. In your example, $\vec{F}$ clearly points in $z$ direction and $\mathrm d\vec{x}$ clearly points in $x$ direction at that instant of time (the electrons move along the wire). Only in the next instant of time, the direction of $\mathrm d\vec{x}$ would have changed, but then the direction of $\vec{F}$ would have changed as well. What you are trying to do is to multiply the Lorentz force at an instant of time with the displacement at a different instant of time, which is wrong.

The equation is $\mathrm d W=\vec{F}\cdot\mathrm d\vec{x}$ and not $\mathrm d W=\vec{F}\cdot\mathrm d\vec{a}$.

13. Aug 15, 2014

Staff: Mentor

Pleco, you are welcome here to learn, not to argue. This is not a debate and it is against the forum rules to make assertions that are contrary to mainstream physics.

14. Aug 15, 2014

Staff: Mentor

No. Both v vectors are the velocity of the charge. Therefore the triple product always gives 0. The math is crystal clear.

15. Aug 15, 2014

pleco

I'm not making any assertions, please ask for reference if you are not familiar with something I said.

This is from Wikipedia:

http://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation

That equation will produce correct results. You can use dx instead of ds only if x stands for "curve" and not x-axis. If you disagree, then please explain your opinion.

16. Aug 15, 2014

rubi

$W=\int F \mathrm d s$ is exactly the same as $W = \int \vec{F}\mathrm d\vec{x}$, where $F(s):=\vec{F}(s)\cdot\vec{v}(s)\frac{\mathrm d t}{\mathrm d s}$, reparametrized in terms of the arclength, so both integrals give you the same number. The second equality $\int F\mathrm d s = F\int \mathrm d s$ is only true if $F(s) = F$ doesn't depend on the arclength at all, which is of course false in your example due to the inhomogeneous magnetic field.

Computing the work using $W=\int \vec{F}\mathrm d\vec{x}$ is completely valid. It is the most general definition of work and all other formulas that you will find are just a special case of it.

The $\vec{x}$ doesn't stand for $x$-axis. It is just the vector $\vec{x} = \begin{pmatrix} x\\y\\z \end{pmatrix}$.

17. Aug 15, 2014

pleco

I'm not sure if we completely understand each other. It can not be a line integral, it must not be limited to change in just one direction. It has to be arbitrary path or curve integral, so electrons may, if they are compelled to do so, move in lateral direction to their initial velocity vector.

18. Aug 16, 2014

rubi

Work is always a line integral. There is no exception. A particle defines a path $C_{t_0}^{t_1} = \{\vec{x}(t) : t_0 \leq t \leq t_1\} \subseteq \mathbb R^3$ in space and a force field is a function that assigns a vector $\vec{F}(\vec{x})$ to each point of space. The work along the path $C_{t_0}^{t_1}$ is defined to be the integral $W = \int_{C_{t_0}^{t_1}} \vec{F}(\vec{x})\,\mathrm d \vec{x}$. This definition captures all occurrences of work in physics. If you want to compute that integral, you must choose some parametrization $\vec{x}(s)$ of the path and compute $\int_{s_0}^{s_1} \vec{F}(\vec{x}(s)) \frac{\mathrm d\vec{x}}{\mathrm d s} \mathrm d s$, but this is not relevant for our discussion, because we can just derive $W=0$ directly from the definition $W=\int \vec{F}\mathrm d\vec{x}$ of work, so let's please stick to that definition.

All of this is completely irrelevant. The only thing we need to know is that the magnetic part of the Lorentz force is always perpendicular to the displacement, so $\vec{F}\cdot\mathrm d \vec{x} = 0$ and thus $W = 0$.

Last edited: Aug 16, 2014
19. Aug 16, 2014

milesyoung

DaleSpam just showed you (in post #9) the solution of the trajectory of the electron in your example. As you can see, the velocity in the x-direction isn't constant, but the magnitude of $\mathbf{v}$ is, i.e. the speed of the electron is constant.

If you single out an electron that has that velocity at some instant in time, then yes (for the magnetic component of the Lorentz force), but it's not the magnetic component of the Lorentz force that does work on the conductor, and no matter how many times you show that image it still won't be a counterexample, and you won't understand why until you get the basics sorted.

And just to be absolutely clear - we're not talking about any conductors. We're just looking at an electron traveling in space.

In this expression:
$$W = \int_{t_1}^{t_2} \mathbf{F_\mathbf{mag}} \cdot \mathbf{v} \, \mathrm{d}t$$
Displacement is not a factor. It uses the definition of work as the integration of the power $\mathbf{F_\mathbf{mag}} \cdot \mathbf{v}$ over the trajectory of the point of application of $\mathbf{F_\mathbf{mag}}$.

Indeed, that's exactly how many define it:
http://en.wikipedia.org/wiki/Work_(physics)#Work_and_potential_energy

You don't even have to muck around with integrals. If the power $\mathbf{F_\mathbf{mag}} \cdot \mathbf{v}$ is always zero, then clearly the magnetic force can't have done any work on the particle. As DaleSpam showed, $\mathbf{F_\mathbf{mag}} \cdot \mathbf{v} = 0$ is indeed always true.

If you insist on talking about displacement (and I've tried to avoid it because you seem to think that displacement is something that must happen in the direction of whatever force is applied):
$$W = F s$$
This is only valid for a constant force that acts in the direction of displacement. Notice that $F$ and $s$ are scalars.

I wonder if this is where you took a wrong turn somewhere. You do realize that the applied force and the displacement don't have to be in the same direction?

This is the general expression:
$$W = \int_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{s}$$
where C is the trajectory of the point of application of $\mathbf{F}$ and $\mathbf{F}$ can vary along C.

But since:
$$W = \int_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{s} = \int_{t_1}^{t_2} \mathbf{F} \cdot \frac{\mathrm{d}\mathbf{s}}{\mathrm{d}t} \, \mathrm{d}t = \int_{t_1}^{t_2} \mathbf{F} \cdot \mathbf{v} \, \mathrm{d}t$$
It changes absolutely nothing.

So to reiterate: You don't get to choose what you substitute into $\mathbf{F_\mathbf{mag}} \cdot \mathbf{v}$. You use the instantaneous magnetic force $\mathbf{F_\mathbf{mag}}$ and the instantaneous velocity $\mathbf{v}$. Period.

Now, this is going to be my last post on this matter. If you want to set up a problem in basic mechanics, like, for instance, uniform circular motion, then I'm all for discussing that instead.

Last edited: Aug 16, 2014
20. Aug 16, 2014

Staff: Mentor

Yes, you are. Besides the post I have deleted you have made several flat out wrong assertions as well as several irrelevant assertions:
Irrelevant assertion.

Wrong assertion. F.v = (0,0,1).(1,0,0) = 0

Wrong assertion, they are never in the same direction with the Lorentz force.

Wrong. The v in both cases is the velocity of the charge.
Wrong assertion. Work is always a line integral.

Irrelevant assertion. Nobody but you is limiting the change to just one direction. Line integrals can be calculated on curved lines and in directions lateral to the initial velocity vector, as I showed earlier in the differential equation.

Irrelevant assertion. If you read the derivation you would see that it is only valid for constant F, which is not the case here.

You need to stop making assertions and start listening to the good advice that you are getting. We want to help you learn, but to do that you need to put forth some effort to understand what is being explained and look to see the many places that your current understanding is flat out wrong.

Last edited: Aug 16, 2014